Prove $\Re\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right) =\frac{1}{2}+\frac{\sin{\left(n+\frac{1}{2}\theta\right)}}{2\sin{\frac{\theta}{2}}}$

Question

With $\text{Re}(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}})$, I expanded it with polar form and found the real component to be

$$\frac{1}{2}+\frac{\cos(n\theta)-\cos((n+1)\theta)}{2{\sin}^2\left(\frac{\theta}{2}\right)}$$

where the numerator was derived from. $(1-\text{cis}(n+1)x) \cdot (1-\cos x+i\sin x)$

I graphed it and it seems different to what is asked.

Have I made a mistake? if not, how do I proceed

Answer 1

Your derivation is correct, now by sum to product formula we obtain

$$\cos(n\theta)-\cos((n+1)\theta) = -2\sin\left(\frac{(2n+1)\theta}2\right)\sin\left(\frac{-\theta}2\right)$$

which leads to the desired expression.